At the moment nobody knows the answer.
What is known can be summed up as follows.

1. Reformulation in the Fourier–analytic language

Put
f := 1_A – 1_B (takes only the values ±1).

The requirement
λ(A∩D)=λ(B∩D)=½λ(D) for every unit Euclidean disc D

is exactly the same as

     (f * χ_D)(x) = 0                  for every x∈ℝ²,        (1)

where χ_D is the indicator of the unit disc and * denotes convolution.
Taking Fourier transforms gives

     \hat f(ξ)\, \hat χ_D(ξ) = 0   for every ξ∈ℝ².      (2)

The Fourier transform of the unit disc is the radial function

     \hat χ_D(ξ) = 2π |ξ|^{-1} J₁(|ξ|),

J₁ being the Bessel-J function of order 1. J₁(|ξ|)≠0 for |ξ| lying in an open neighbourhood of 0 and its zero–set is the countable union

     Z := ⋃_{k≥1} {|ξ| = ρ_k },              J₁(ρ_k)=0.

Hence (2) forces

     supp \hat f ⊂ Z.                                 (3)

Thus any kaleidoscopic colouring would have to have all of its “frequency spectrum’’ sitting on the above family of concentric circles.

2. A bounded solution exists if one does not insist on the values being exactly ±1.
   In fact, already a single plane wave works:

    f(x)=cos(ρ₁ ω·x)              (ω any unit vector, ρ₁ the first zero of J₁).
   

Its Fourier transform is supported at the two points ±ρ₁ω, which belong to Z, so (3) holds and hence (1) holds. By taking linear combinations of such waves (or suitable random series supported in Z) one can manufacture bounded measurable functions – with ‖f‖_\infty arbitrarily close to 1 – that satisfy (1). What is missing is the step from “|f|≤1’’ to “|f|=1 almost everywhere’’.

3. Why it is hard to force the values to be exactly ±1.

• The set Z has Lebesgue measure 0 and Hausdorff dimension 1.
A basic theorem of harmonic analysis says that an L¹–function cannot have its Fourier transform supported on such a thin set, so condition (3) already prevents any L¹–solution. When one only assumes f∈L^∞ this obstruction disappears, but then (as soon as one leaves the class of exponentials and trigonometric polynomials) very little is known about what the spectrum can look like.

• If, on top of (3), one demands that the spatial function takes only the two values ±1, one encounters a severe rigidity problem. No general machinery is presently available that would allow one to decide whether a distribution whose Fourier support is contained in a 1-dimensional submanifold can arise from such a “two–level’’ function.

4. Negative results that are known

• Imposing any regularity on the pieces A,B beyond measurability quickly makes (1) impossible.
– If A or B is assumed to be open, or to have locally finite perimeter, the convolution with χ_D cannot vanish identically (the proof uses the structure of the boundary and the fact that χ_D has a jump at |x|=1).
– No periodic colouring with finitely many boundary components in a fundamental domain can work (the Fourier coefficients then decay too fast to live on the set Z).

• Replacing the Euclidean norm by the ℓ_1- or ℓ_∞-norm makes the problem easy: a “chess–board’’ colouring with suitably chosen step size does the job, because the Fourier transform of the corresponding diamond / square vanishes on a whole cross of lines, so that a periodic ±1-function can be arranged whose spectrum sits inside those lines.

5. What would settle the question

One needs either

 (A)  to construct a ±1-valued measurable function whose Fourier transform (as a tempered distribution) is supported on Z, or  

 (B)  to prove that **no** bounded two–valued function can have its spectrum contained in a set of Hausdorff dimension 1 in the plane.  

Both directions are open; no argument presently manages to bridge the gap between the partial positive result of 2. and the partial negative results of 4.

To summarise:

• Bounded measurable solutions of (1) exist, but they are not known to take only the values ±1.
• All available structural information points to a serious obstacle, yet there is no proof that the obstacle is insurmountable.
• Consequently, the existence of a “kaleidoscopic’’ two–colouring of the Euclidean plane remains an open problem.